Solve, using cross - multiplication :
4x - 3y - 11 = 0
6x + 7y - 5 = 0

Answer

To solve the system of equations using cross-multiplication, we start with the given equations:

1. \( 4x - 3y - 11 = 0 \)
2. \( 6x + 7y - 5 = 0 \)

### Step 1: Rewrite the equations in standard form
We can rewrite the equations in the form \( ax + by + c = 0 \):

- For the first equation:
\[ 4x - 3y = 11 \quad \text{(Equation 1)} \]
- For the second equation:
\[ 6x + 7y = 5 \quad \text{(Equation 2)} \]

### Step 2: Identify coefficients
From the equations, we identify the coefficients:
- For Equation 1: \( a_1 = 4, b_1 = -3, c_1 = -11 \)
- For Equation 2: \( a_2 = 6, b_2 = 7, c_2 = -5 \)

### Step 3: Apply the cross-multiplication formula
The formulas for \( x \) and \( y \) using cross-multiplication are:
\[ x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1} \]
\[ y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1} \]

### Step 4: Substitute the values into the formulas
Substituting the identified coefficients into the formulas:

For \( x \):
\[ x = \frac{(-3)(-5) - (7)(-11)}{(4)(7) - (6)(-3)} \]

For \( y \):
\[ y = \frac{(-11)(6) - (-5)(4)}{(4)(7) - (6)(-3)} \]

### Step 5: Calculate the numerator and denominator for \( x \)
Calculating the numerator:
\[ (-3)(-5) = 15 \]
\[ (7)(-11) = -77 \]
So,
\[ b_1 c_2 - b_2 c_1 = 15 + 77 = 92 \]

Calculating the denominator:
\[ (4)(7) = 28 \]
\[ (6)(-3) = -18 \]
So,
\[ a_1 b_2 - a_2 b_1 = 28 + 18 = 46 \]

Thus, we have:
\[ x = \frac{92}{46} = 2 \]

### Step 6: Calculate the numerator and denominator for \( y \)
Calculating the numerator:
\[ (-11)(6) = -66 \]
\[ (-5)(4) = -20 \]
So,
\[ c_1 a_2 - c_2 a_1 = -66 + 20 = -46 \]

Using the same denominator as before:
\[ a_1 b_2 - a_2 b_1 = 46 \]

Thus, we have:
\[ y = \frac{-46}{46} = -1 \]

### Final Solution
The solution to the system of equations is:
\[ x = 2, \quad y = -1 \]